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ollegr [7]
2 years ago
15

Help please I’ve been here for a while...

Mathematics
2 answers:
Minchanka [31]2 years ago
6 0
B. should be correct
lukranit [14]2 years ago
4 0
B just trust me bro I got you
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The length of a rectangle is 3cm more than twice the width. The perimeter of the rectangle is 42 cm. Find the dimensions of the
bogdanovich [222]
The width is 6cm
And the length is two times + 3 of the width
6 X 2 + 3 = 15
Add all 4 sides for the perimeter 6+6+15+15=42
The the dimensions are a rectangle that is 6cm wide and 15cm tall
5 0
3 years ago
#1) What is the most efficient way to isolate and solve for the variable? *
s344n2d4d5 [400]
Subtract 2x from both sides
6 0
3 years ago
Read 2 more answers
A collection of stamps consists of 3c stamps 5c stamps and 7c stamps. There are 6 more 3c stamps than 5c stamps and two more 7c
Deffense [45]

The number of 3¢ stamps that are in the collection is: 22.

<h3>What is an expression?</h3>

An expression can be defined as a mathematical equation which is used to show the relationship existing between two or more variables and numerical quantities.

In this exercise, you're required to determine the number of 3 cents (3¢) . Therefore, we would translate the word problem into an algebraic expression and then evaluate it as follows:

<u>Note:</u> 100 cents (¢) is equal to 1 dollar ($).

A collection of stamps consists of 3¢ stamps, 5¢ stamps and 7¢ stamps; 3¢ + 5¢ + 7¢ = 194

6 more 3¢ stamps than 5¢ stamps; 3¢ = 5¢ + 6  ⇒ 5¢ = 3¢ - 6

Two more 7¢ stamps than 3¢ stamps; 7¢ = 3¢ + 2

Combining the above equations, we have:

3¢ + 3¢ - 6 + 3¢ + 2 = 194

9¢ = 198

¢ = 198/9

¢ = 22.

Read more on expressions here: brainly.com/question/17361494

#SPJ1

<u>Complete Question:</u>

A collection of stamps consists of 3¢ stamps 5¢ stamps and 7¢ stamps. There are 6 more 3¢ stamps than 5¢ stamps and two more 7¢ stamps than 3¢ stamps. The total value of the stamps is $1.94. How many 3¢ stamps are in the collection?

6 0
1 year ago
I need help with a math question
Julli [10]

Answer:

Question 1: 4.5(10^−7)(2(10^4))

=

9/1000

(Decimal: 0.009)

Question 2:

mass of neutron/mass of electron = 2*10-24/(9*10-28)

2x10^(-24)/9x10^(-28)

d is closest.

Question 3:

4(10^3)(12(10^5))

=4000*1200000

=4*1000*12*100000

=(4*12)*(1000*100000)

=48*100000000

=4800000000

And these things : ^ mean raising the number to become exponets and when I put the number into a bold text thats the answer!

<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>!</em><em> </em><em>if</em><em> </em><em>so</em><em> </em><em>pls</em><em> </em><em>mark</em><em> </em><em>brainliest</em><em> </em><em>and</em><em> </em><em>heart</em><em>/</em><em>rate</em><em>!</em>

3 0
3 years ago
Calculate the total installment price the carrying charges in the number of months needed to save them money at the monthly rate
masha68 [24]

Answer:

  • the total installment is $936
  • the carrying charges are $186
  • the monthly rate to buy the items are  19.23 or 20 months

Step-by-step explanation:

$39 for 24 months comes to ...

$39 × 24 = $936

The "carrying charges" are the difference between the cash price and the installment price:

$936 -750 = $186

 Saving at the rate of $39 per month, it would take ...

  $750/$39 = 19.23 months   about 20 months to save up the cash price.

Often, such savings would come from a once-a-month paycheck, so it would take more than 19 such paychecks to save the required amount.

hope it will help :)

3 0
3 years ago
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